Question

A torque of 60 N*m acts on a wheel of moment of inertia 30 kg*m^2 for 5 s and then is removed.
a) What is the angular acceleration of the wheel?
b) How many revolutions does it make in 15 s if it starts at rest?

Answer 1

\[\tau=I a\]

Answer 2

\[60=30 * a\]

Answer 3

k thx and wat about b) which formula do i apply?

Answer 4

You have the angular acceleration which is:\[\alpha =d^2\theta/dt^2\]so, to figure out how many revolutions have happened in 15s we integrate twice to get the total angle swept out:\[d^2\theta=\alpha dt\]and,\[\Delta \theta=\int\limits_{0}^{15}2dt=30\]One revolution represents\[\Delta \theta=2\pi\]so 30 radians represents:\[\frac{30}{2\pi}\approx4.77\]So about 4.77 revolutions have occurred in 15s, starting from rest.

Answer 5

ah crap I only integrated once here. We need to integrate twice! lol

Answer 6

\[\Delta \theta=15^2=225\]So we get\[\frac{225}{2\pi}\approx35.8\]so 35.8 revolutions should be correct :)

Answer 7

i tried 35.8 but apparently its wrong

Answer 8

5sec , not 15 sec

Answer 9

ooops...the force only acts for 5 seconds not 15! so for 10 seconds the angular acceleration is zero (for the first 5s the angular acceleration is 2)

Answer 10

gotta go tho...this should be enough for you to solve it :)

Answer 11

ok, in case you still needed help on this one:\[\theta=\frac{1}{2}\alpha t^2+\omega t+\theta _{0}\]This is the general equation of motion you need. Setting initial angular position=0 and angular acceleration=2:\[\theta=5^2+2(10)+0=45\]So total number of revolutions is:\[\frac{45}{2\pi}\approx 7.2\]

Answer 12

i got 19.894

Answer 13

yeah I think I see my error

Answer 14

5^2+2(5)(10)+0=125
and 125/2pi=19.9

Answer 15

good call

Answer 16

thank you though

Answer 17

np...im a little rusty hehe